WEBVTT 00:00:12.079 --> 00:00:15.760 This guy, Grothendieck, is somewhat of a mathematical idol to me, 00:00:15.759 --> 00:00:17.879 and I just love this quote, don't you? 00:00:18.620 --> 00:00:22.379 Too often in math, we dive into showing that a certain fact is true 00:00:22.379 --> 00:00:27.298 with a long series of formulas before stepping back and making sure it feels reasonable, 00:00:27.298 --> 00:00:30.339 and preferably obvious, at least at an intuitive level. 00:00:31.260 --> 00:00:33.792 In this video, I want to talk about integrals, 00:00:33.792 --> 00:00:37.619 and the thing that I want to become almost obvious is that they are an 00:00:37.619 --> 00:00:38.859 inverse of derivatives. 00:00:39.899 --> 00:00:42.125 Here we're just going to focus on one example, 00:00:42.125 --> 00:00:46.198 which is a kind of dual to the example of a moving car that I talked about in chapter 00:00:46.198 --> 00:00:48.140 2 of the series, introducing derivatives. 00:00:49.179 --> 00:00:52.618 Then in the next video we're going to see how this same idea generalizes, 00:00:52.618 --> 00:00:54.060 but to a couple other contexts. 00:00:55.240 --> 00:00:58.849 Imagine you're sitting in a car, and you can't see out the window, 00:00:58.848 --> 00:01:00.519 all you see is the speedometer. 00:01:02.079 --> 00:01:05.632 At some point the car starts moving, speeds up, 00:01:05.632 --> 00:01:10.740 and then slows back down to a stop, all over the course of 8 seconds. 00:01:11.680 --> 00:01:15.212 The question is, is there a nice way to figure out how far you've 00:01:15.212 --> 00:01:18.960 travelled during that time based only on your view of the speedometer? 00:01:19.539 --> 00:01:23.028 Or better yet, can you find a distance function, s of t, 00:01:23.028 --> 00:01:27.497 that tells you how far you've travelled after a given amount of time, t, 00:01:27.498 --> 00:01:29.579 somewhere between 0 and 8 seconds? 00:01:30.900 --> 00:01:34.000 Let's say you take note of the velocity at every second, 00:01:34.000 --> 00:01:37.099 and make a plot over time that looks something like this. 00:01:38.959 --> 00:01:43.149 And maybe you find that a nice function to model that velocity 00:01:43.150 --> 00:01:47.540 over time in meters per second is v of t equals t times 8 minus t. 00:01:48.418 --> 00:01:51.786 You might remember, in chapter 2 of this series we were looking at 00:01:51.786 --> 00:01:55.253 the opposite situation, where you knew what a distance function was, 00:01:55.253 --> 00:01:58.719 s of t, and you wanted to figure out the velocity function from that. 00:01:59.739 --> 00:02:02.500 There I showed how the derivative of a distance vs. 00:02:02.560 --> 00:02:04.859 time function gives you a velocity vs. 00:02:04.920 --> 00:02:05.600 time function. 00:02:06.359 --> 00:02:09.615 So in our current situation, where all we know is velocity, 00:02:09.615 --> 00:02:12.219 it should make sense that finding a distance vs. 00:02:12.280 --> 00:02:15.432 time function is going to come down to asking what 00:02:15.432 --> 00:02:18.340 function has a derivative of t times 8 minus t. 00:02:19.379 --> 00:02:23.364 This is often described as finding the antiderivative of a function, and indeed, 00:02:23.364 --> 00:02:27.300 that's what we'll end up doing, and you could even pause right now and try that. 00:02:27.900 --> 00:02:32.641 But first, I want to spend the bulk of this video showing how this question is related 00:02:32.640 --> 00:02:35.420 to finding the area bounded by the velocity graph, 00:02:35.420 --> 00:02:39.344 because that helps to build an intuition for a whole class of problems, 00:02:39.344 --> 00:02:42.180 things called integral problems in math and science. 00:02:42.780 --> 00:02:45.884 To start off, notice that this question would be a lot easier 00:02:45.884 --> 00:02:48.739 if the car was just moving at a constant velocity, right? 00:02:49.419 --> 00:02:54.152 In that case, you could just multiply the velocity in meters per second times the amount 00:02:54.152 --> 00:02:58.939 of time that has passed in seconds, and that would give you the number of meters traveled. 00:03:00.020 --> 00:03:04.159 And notice, you can visualize that product, that distance, as an area. 00:03:05.000 --> 00:03:08.819 And if visualizing distance as area seems kind of weird, I'm right there with you. 00:03:08.819 --> 00:03:13.847 It's just that on this plot, where the horizontal direction has units of seconds, 00:03:13.848 --> 00:03:17.467 and the vertical direction has units of meters per second, 00:03:17.467 --> 00:03:20.840 units of area just very naturally correspond to meters. 00:03:22.020 --> 00:03:25.742 But what makes our situation hard is that velocity is not constant, 00:03:25.741 --> 00:03:28.479 it's incessantly changing at every single instant. 00:03:30.780 --> 00:03:35.230 It would even be a lot easier if it only ever changed at a handful of points, 00:03:35.230 --> 00:03:39.626 maybe staying static for the first second, and then suddenly discontinuously 00:03:39.626 --> 00:03:43.849 jumping to a constant 7 meters per second for the next second, and so on, 00:03:43.848 --> 00:03:47.159 with discontinuous jumps to portions of constant velocity. 00:03:48.699 --> 00:03:51.208 That would make it uncomfortable for the driver, 00:03:51.209 --> 00:03:55.356 in fact it's actually physically impossible, but it would make your calculations 00:03:55.356 --> 00:03:56.740 a lot more straightforward. 00:03:57.599 --> 00:04:02.957 You could just compute the distance traveled on each interval by multiplying the constant 00:04:02.957 --> 00:04:07.900 velocity on that interval by the change in time, and then just add all of those up. 00:04:09.020 --> 00:04:13.075 So what we're going to do is approximate the velocity function as if it 00:04:13.074 --> 00:04:17.242 was constant on a bunch of intervals, and then, as is common in calculus, 00:04:17.242 --> 00:04:21.579 we'll see how refining that approximation leads us to something more precise. 00:04:24.720 --> 00:04:27.740 Here, let's make this a little more concrete by throwing in some numbers. 00:04:28.360 --> 00:04:33.673 Chop up the time axis between 0 and 8 seconds into many small intervals, 00:04:33.673 --> 00:04:38.040 each with some little width dt, something like 0.25 seconds. 00:04:38.939 --> 00:04:43.920 Consider one of those intervals, like the one between t equals 1 and 1.25. 00:04:45.279 --> 00:04:49.681 In reality, the car speeds up from 7 m per second to about 8.4 m per 00:04:49.682 --> 00:04:53.893 second during that time, and you could find those numbers just by 00:04:53.892 --> 00:04:58.359 plugging in t equals 1 and t equals 1.25 to the equation for velocity. 00:04:59.459 --> 00:05:02.070 What we want to do is approximate the car's motion 00:05:02.071 --> 00:05:04.580 as if its velocity was constant on that interval. 00:05:05.540 --> 00:05:08.566 Again, the reason for doing that is we don't really know 00:05:08.565 --> 00:05:11.699 how to handle situations other than constant velocity ones. 00:05:13.459 --> 00:05:17.719 You could choose this constant to be anything between 7 and 8.4. 00:05:18.019 --> 00:05:19.240 It actually doesn't matter. 00:05:20.019 --> 00:05:23.536 All that matters is that our sequence of approximations, 00:05:23.536 --> 00:05:28.040 whatever they are, gets better and better as dt gets smaller and smaller. 00:05:28.740 --> 00:05:32.485 That treating this car's journey as a bunch of discontinuous jumps 00:05:32.485 --> 00:05:36.288 in speed between portions of constant velocity becomes a less-wrong 00:05:36.288 --> 00:05:39.979 reflection of reality as we decrease the time between those jumps. 00:05:42.540 --> 00:05:46.555 So for convenience, on an interval like this, let's just approximate the 00:05:46.555 --> 00:05:50.845 speed with whatever the true car's velocity is at the start of that interval, 00:05:50.845 --> 00:05:54.640 the height of the graph above the left side, which in this case is 7. 00:05:55.959 --> 00:06:00.117 In this example interval, according to our approximation, 00:06:00.117 --> 00:06:03.560 the car moves 7 m per second times 0.25 seconds. 00:06:04.459 --> 00:06:09.779 That's 1.75 meters, and it's nicely visualized as the area of this thin rectangle. 00:06:10.699 --> 00:06:14.439 In truth, that's a little under the real distance traveled, but not by much. 00:06:14.980 --> 00:06:16.920 The same goes for every other interval. 00:06:17.420 --> 00:06:22.479 The approximated distance is v of t times dt, it's just that you'd be plugging in a 00:06:22.478 --> 00:06:27.839 different value for t at each one of these, giving a different height for each rectangle. 00:06:29.959 --> 00:06:32.525 I'm going to write out an expression for the sum of 00:06:32.526 --> 00:06:35.340 the areas of all those rectangles in kind of a funny way. 00:06:36.019 --> 00:06:40.120 Take this symbol here, which looks like a stretched s for sum, 00:06:40.120 --> 00:06:43.180 and put a 0 at its bottom and an 8 at its top, 00:06:43.180 --> 00:06:48.000 to indicate that we'll be ranging over time steps between 0 and 8 seconds. 00:06:48.899 --> 00:06:54.399 And as I said, the amount we're adding up at each time step is v of t times dt. 00:06:55.459 --> 00:06:57.459 Two things are implicit in this notation. 00:06:58.180 --> 00:07:01.340 First of all, that value dt plays two separate roles. 00:07:01.920 --> 00:07:05.263 Not only is it a factor in each quantity we're adding up, 00:07:05.262 --> 00:07:08.779 it also indicates the spacing between each sampled time step. 00:07:09.379 --> 00:07:13.464 So when you make dt smaller and smaller, even though it decreases the area of 00:07:13.464 --> 00:07:18.125 each rectangle, it increases the total number of rectangles whose areas we're adding up, 00:07:18.125 --> 00:07:21.739 because if they're thinner, it takes more of them to fill that space. 00:07:22.879 --> 00:07:28.100 And second, the reason we don't use the usual sigma notation to indicate a sum is that 00:07:28.100 --> 00:07:33.260 this expression is technically not any particular sum for any particular choice of dt. 00:07:33.779 --> 00:07:38.419 It's meant to express whatever that sum approaches as dt approaches 0. 00:07:39.480 --> 00:07:42.379 And as you can see, what that approaches is the 00:07:42.379 --> 00:07:45.460 area bounded by this curve and the horizontal axis. 00:07:46.339 --> 00:07:51.753 Remember, smaller choices of dt indicate closer approximations for the original question, 00:07:51.754 --> 00:07:53.740 how far does the car actually go? 00:07:54.540 --> 00:07:58.463 So this limiting value for the sum, the area under this curve, 00:07:58.463 --> 00:08:03.259 gives us the precise answer to the question in full unapproximated precision. 00:08:04.300 --> 00:08:05.540 Now tell me that's not surprising. 00:08:06.060 --> 00:08:09.497 We had this pretty complicated idea of approximations that 00:08:09.497 --> 00:08:12.759 can involve adding up a huge number of very tiny things. 00:08:13.480 --> 00:08:18.200 And yet, the value that those approximations approach can be described so simply, 00:08:18.199 --> 00:08:20.560 it's just the area underneath this curve. 00:08:22.120 --> 00:08:25.112 This expression is called an integral of v of t, 00:08:25.112 --> 00:08:28.960 since it brings all of its values together, it integrates them. 00:08:30.060 --> 00:08:32.820 Now at this point, you could say, how does this help? 00:08:33.240 --> 00:08:37.510 You've just reframed one hard question, finding how far the car has traveled, 00:08:37.510 --> 00:08:42.439 into an equally hard problem, finding the area between this graph and the horizontal axis. 00:08:43.879 --> 00:08:44.779 And you'd be right. 00:08:45.259 --> 00:08:50.254 If the velocity-distance duo was the only thing we cared about, most of this video, 00:08:50.254 --> 00:08:54.240 with all the area under a curve nonsense, would be a waste of time. 00:08:54.659 --> 00:08:57.259 We could just skip straight ahead to finding an antiderivative. 00:08:58.000 --> 00:09:02.351 But finding the area between a function's graph and the horizontal axis 00:09:02.351 --> 00:09:06.765 is somewhat of a common language for many disparate problems that can be 00:09:06.765 --> 00:09:11.240 broken down and approximated as the sum of a large number of small things. 00:09:12.340 --> 00:09:15.836 You'll see more in the next video, but for now I'll just say in 00:09:15.836 --> 00:09:19.115 the abstract that understanding how to interpret and how to 00:09:19.115 --> 00:09:22.940 compute the area under a graph is a very general problem-solving tool. 00:09:23.600 --> 00:09:28.094 In fact, the first video of this series already covered the basics of how this works, 00:09:28.094 --> 00:09:31.229 but now that we have more of a background with derivatives, 00:09:31.229 --> 00:09:33.320 we can take this idea to its completion. 00:09:34.320 --> 00:09:39.580 For a velocity example, think of this right endpoint as a variable, capital T. 00:09:41.679 --> 00:09:45.829 So we're thinking of this integral of the velocity function between 0 and T, 00:09:45.830 --> 00:09:48.417 the area under this curve between those inputs, 00:09:48.417 --> 00:09:51.220 as a function where the upper bound is the variable. 00:09:52.059 --> 00:09:56.899 That area represents the distance the car has travelled after T seconds, right? 00:09:57.379 --> 00:09:59.299 So in reality, this is a distance vs. 00:09:59.360 --> 00:10:01.279 time function, s of t. 00:10:01.899 --> 00:10:04.819 Now ask yourself, what is the derivative of that function? 00:10:07.299 --> 00:10:12.323 On the one hand, a tiny change in distance over a tiny change in time is velocity, 00:10:12.323 --> 00:10:14.019 that is what velocity means. 00:10:14.840 --> 00:10:19.139 But there's another way to see this, purely in terms of this graph and this area, 00:10:19.139 --> 00:10:22.180 which generalizes a lot better to other integral problems. 00:10:23.299 --> 00:10:27.854 A slight nudge of dt to the input causes that area to increase, 00:10:27.855 --> 00:10:31.700 some little ds represented by the area of this sliver. 00:10:32.740 --> 00:10:37.130 The height of that sliver is the height of the graph at that point, 00:10:37.130 --> 00:10:38.939 v of t, and its width is dt. 00:10:39.779 --> 00:10:45.297 And for small enough dt, we can basically consider that sliver to be a rectangle, 00:10:45.297 --> 00:10:50.680 so this little bit of added area, ds, is approximately equal to v of t times dt. 00:10:51.659 --> 00:10:56.967 And because that's an approximation that gets better and better for smaller dt, 00:10:56.967 --> 00:11:01.679 the derivative of that area function, ds, dt, at this point equals vt, 00:11:01.679 --> 00:11:06.059 the value of the velocity function at whatever time we started on. 00:11:06.980 --> 00:11:09.259 And that right there is a super general argument. 00:11:09.259 --> 00:11:12.682 The derivative of any function giving the area under a 00:11:12.682 --> 00:11:16.539 graph like this is equal to the function for the graph itself. 00:11:18.740 --> 00:11:24.440 So, if our velocity function is t times 8-t, what should s be? 00:11:25.139 --> 00:11:28.699 What function of t has a derivative of t times 8-t? 00:11:30.340 --> 00:11:34.767 It's easier to see if we expand this out, writing it as 8t minus t squared, 00:11:34.767 --> 00:11:37.680 and then we can just take each part one at a time. 00:11:37.679 --> 00:11:40.919 What function has a derivative of 8t? 00:11:42.240 --> 00:11:45.570 We know that the derivative of t squared is 2t, 00:11:45.570 --> 00:11:50.912 so if we just scale that up by a factor of 4, we can see that the derivative 00:11:50.912 --> 00:11:52.300 of 4t squared is 8t. 00:11:53.019 --> 00:11:55.659 And for that second part, what kind of function do 00:11:55.659 --> 00:11:58.559 you think might have negative t squared as a derivative? 00:12:00.200 --> 00:12:04.929 Using the power rule again, we know that the derivative of a cubic term, 00:12:04.928 --> 00:12:07.779 t cubed, gives us a square term, 3t squared. 00:12:08.480 --> 00:12:11.201 So if we just scale that down by a third, the 00:12:11.201 --> 00:12:14.220 derivative of 1 third t cubed is exactly t squared. 00:12:14.919 --> 00:12:17.942 And then making that negative, we'd see that negative 00:12:17.942 --> 00:12:21.019 1 third t cubed has a derivative of negative t squared. 00:12:22.179 --> 00:12:26.168 Therefore, the antiderivative of our function, 00:12:26.168 --> 00:12:30.919 8t minus t squared, is 4t squared minus 1 third t cubed. 00:12:32.438 --> 00:12:34.159 But there's a slight issue here. 00:12:34.480 --> 00:12:37.914 We could add any constant we want to this function, 00:12:37.914 --> 00:12:41.019 and its derivative is still 8t minus t squared. 00:12:41.820 --> 00:12:44.500 The derivative of a constant always goes to zero. 00:12:45.179 --> 00:12:49.375 And if you were to graph s of t, you could think of this in the sense that moving a 00:12:49.375 --> 00:12:53.820 graph of a distance function up and down does nothing to affect its slope at every input. 00:12:54.639 --> 00:12:58.640 So in reality, there's actually infinitely many different 00:12:58.640 --> 00:13:02.986 possible antiderivative functions, and every one of them looks 00:13:02.986 --> 00:13:07.539 like 4t squared minus 1 third t cubed plus c, for some constant c. 00:13:08.580 --> 00:13:12.782 But there is one piece of information we haven't used yet that will let 00:13:12.782 --> 00:13:17.160 us zero in on which antiderivative to use, the lower bound of the integral. 00:13:18.360 --> 00:13:21.403 This integral has to be zero when we drag that right 00:13:21.403 --> 00:13:24.220 endpoint all the way to the left endpoint, right? 00:13:24.639 --> 00:13:30.379 The distance travelled by the car between 0 seconds and 0 seconds is… well, zero. 00:13:31.580 --> 00:13:34.845 So as we found, the area as a function of capital 00:13:34.845 --> 00:13:37.720 T is an antiderivative for the stuff inside. 00:13:38.480 --> 00:13:42.069 And to choose what constant to add to this expression, 00:13:42.068 --> 00:13:47.159 you subtract off the value of that antiderivative function at the lower bound. 00:13:48.159 --> 00:13:51.815 If you think about it for a moment, that ensures that the 00:13:51.816 --> 00:13:55.600 integral from the lower bound to itself will indeed be zero. 00:13:57.740 --> 00:14:02.471 As it so happens, when you evaluate the function we have here at t equals zero, 00:14:02.471 --> 00:14:03.240 you get zero. 00:14:03.919 --> 00:14:07.219 So in this specific case, you don't need to subtract anything off. 00:14:07.980 --> 00:14:13.500 For example, the total distance travelled during the full 8 seconds 00:14:13.500 --> 00:14:18.940 is this expression evaluated at t equals 8, which is 85.33 minus 0. 00:14:18.940 --> 00:14:22.060 So the answer as a whole is 85.33. 00:14:23.179 --> 00:14:27.459 But a more typical example would be something like the integral between 1 and 7. 00:14:28.200 --> 00:14:30.968 That's the area pictured here, and it represents 00:14:30.967 --> 00:14:34.019 the distance travelled between 1 second and 7 seconds. 00:14:36.480 --> 00:14:41.146 What you do is evaluate the antiderivative we found at the top bound, 00:14:41.145 --> 00:14:44.679 7, and subtract off its value at the bottom bound, 1. 00:14:45.899 --> 00:14:50.159 Notice, by the way, it doesn't matter which antiderivative we chose here. 00:14:50.559 --> 00:14:56.559 If for some reason it had a constant added to it, like 5, that constant would cancel out. 00:14:58.000 --> 00:15:03.080 More generally, any time you want to integrate some function, and remember, 00:15:03.080 --> 00:15:08.762 you think of that as adding up values f of x times dx for inputs in a certain range, 00:15:08.761 --> 00:15:12.839 and then asking what is that sum approach as dx approaches 0. 00:15:13.659 --> 00:15:18.312 The first step to evaluating that integral is to find an antiderivative, 00:15:18.312 --> 00:15:23.539 some other function, capital F, whose derivative is the thing inside the integral. 00:15:24.799 --> 00:15:28.435 Then the integral equals this antiderivative evaluated 00:15:28.436 --> 00:15:31.940 at the top bound minus its value at the bottom bound. 00:15:32.820 --> 00:15:37.460 And this fact right here that you're staring at is the fundamental theorem of calculus. 00:15:38.240 --> 00:15:41.259 And I want you to appreciate something kind of crazy about this fact. 00:15:41.840 --> 00:15:46.234 The integral, the limiting value for the sum of all these thin rectangles, 00:15:46.234 --> 00:15:49.516 takes into account every single input on the continuum, 00:15:49.515 --> 00:15:51.860 from the lower bound to the upper bound. 00:15:52.279 --> 00:15:55.839 That's why we use the word integrate, it brings them all together. 00:15:56.879 --> 00:16:00.506 And yet, to actually compute it using an antiderivative, 00:16:00.506 --> 00:16:04.579 you only look at two inputs, the top bound and the bottom bound. 00:16:05.419 --> 00:16:06.559 It almost feels like cheating. 00:16:06.940 --> 00:16:10.841 Finding the antiderivative implicitly accounts for all the 00:16:10.841 --> 00:16:15.139 information needed to add up the values between those two bounds. 00:16:15.919 --> 00:16:17.339 That's just crazy to me. 00:16:18.679 --> 00:16:22.418 This idea is deep, and there's a lot packed into this whole concept, 00:16:22.418 --> 00:16:25.399 so let's recap everything that just happened, shall we? 00:16:26.220 --> 00:16:30.580 We wanted to figure out how far a car goes just by looking at the speedometer. 00:16:31.360 --> 00:16:34.220 And what makes that hard is that velocity is always changing. 00:16:35.078 --> 00:16:39.351 If you approximate velocity to be constant on multiple different intervals, 00:16:39.351 --> 00:16:43.848 you could figure out how far the car goes on each interval with multiplication, 00:16:43.849 --> 00:16:45.480 and then add all of those up. 00:16:46.440 --> 00:16:50.639 Better and better approximations for the original problem correspond to 00:16:50.639 --> 00:16:54.779 collections of rectangles whose aggregate area is closer and closer to 00:16:54.779 --> 00:16:58.980 being the area under this curve between the start time and the end time. 00:16:58.980 --> 00:17:03.230 So that area under the curve is actually the precise distance 00:17:03.230 --> 00:17:07.140 traveled for the true nowhere constant velocity function. 00:17:08.400 --> 00:17:11.546 If you think of that area as a function itself, 00:17:11.546 --> 00:17:15.939 with a variable right endpoint, you can deduce that the derivative 00:17:15.939 --> 00:17:20.660 of that area function must equal the height of the graph at every point. 00:17:21.358 --> 00:17:22.759 And that's really the key right there. 00:17:22.759 --> 00:17:26.112 It means that to find a function giving this area, 00:17:26.112 --> 00:17:29.400 you ask, what function has v of t as a derivative? 00:17:30.640 --> 00:17:34.496 There are actually infinitely many antiderivatives of a given function, 00:17:34.496 --> 00:17:38.673 since you can always just add some constant without affecting the derivative, 00:17:38.673 --> 00:17:43.439 so you account for that by subtracting off the value of whatever antiderivative function 00:17:43.439 --> 00:17:45.100 you choose at the bottom bound. 00:17:46.259 --> 00:17:51.980 By the way, one important thing to bring up before we leave is the idea of negative area. 00:17:53.039 --> 00:17:57.539 What if the velocity function was negative at some point, meaning the car goes backwards? 00:17:58.660 --> 00:18:03.340 It's still true that a tiny distance traveled ds on a little time interval is 00:18:03.339 --> 00:18:08.079 about equal to the velocity at that time multiplied by the tiny change in time. 00:18:08.640 --> 00:18:13.072 It's just that the number you'd plug in for velocity would be negative, 00:18:13.071 --> 00:18:15.719 so the tiny change in distance is negative. 00:18:16.799 --> 00:18:21.522 In terms of our thin rectangles, if a rectangle goes below the horizontal axis, 00:18:21.522 --> 00:18:25.596 like this, its area represents a bit of distance traveled backwards, 00:18:25.596 --> 00:18:29.788 so if what you want in the end is to find a distance between the car's 00:18:29.788 --> 00:18:34.099 start point and its end point, this is something you'll want to subtract. 00:18:35.059 --> 00:18:36.839 And that's generally true of integrals. 00:18:37.359 --> 00:18:40.045 Whenever a graph dips below the horizontal axis, 00:18:40.046 --> 00:18:44.980 the area between that portion of the graph and the horizontal axis is counted as negative. 00:18:46.000 --> 00:18:50.089 What you'll commonly hear is that integrals don't measure area per se, 00:18:50.089 --> 00:18:54.179 they measure the signed area between the graph and the horizontal axis. 00:18:55.680 --> 00:18:58.689 Next up, I'm going to bring up more context where this idea 00:18:58.689 --> 00:19:01.047 of an integral and area under curves comes up, 00:19:01.047 --> 00:19:04.759 along with some other intuitions for this fundamental theorem of calculus. 00:19:06.480 --> 00:19:10.363 Maybe you remember, chapter 2 of this series introducing the derivative 00:19:10.363 --> 00:19:13.815 was sponsored by The Art of Problem Solving, so I think there's 00:19:13.815 --> 00:19:18.345 something elegant to the fact that this video, which is kind of a duel to that one, 00:19:18.345 --> 00:19:21.420 was also supported in part by The Art of Problem Solving. 00:19:22.160 --> 00:19:25.189 I really can't imagine a better sponsor for this channel, 00:19:25.189 --> 00:19:29.160 because it's a company whose books and courses I recommend to people anyway. 00:19:29.759 --> 00:19:33.849 They were highly influential to me when I was a student developing a love for 00:19:33.849 --> 00:19:37.886 creative math, so if you're a parent looking to foster your own child's love 00:19:37.886 --> 00:19:42.081 for the subject, or if you're a student who wants to see what math has to offer 00:19:42.082 --> 00:19:46.120 beyond rote schoolwork, I cannot recommend The Art of Problem Solving enough. 00:19:46.740 --> 00:19:50.928 Whether that's their newest development to build the right intuitions in 00:19:50.928 --> 00:19:55.461 elementary school kids, called Beast Academy, or their courses in higher-level 00:19:55.461 --> 00:19:59.420 topics and contest preparation, going to aops.com slash 3blue1brown, 00:19:59.420 --> 00:20:04.412 or clicking on the link in the description, lets them know you came from this channel, 00:20:04.412 --> 00:20:08.200 which may encourage them to support future projects like this one. 00:20:08.920 --> 00:20:13.673 I consider these videos a success not when they teach people a particular bit of math, 00:20:13.673 --> 00:20:17.769 which can only ever be a drop in the ocean, but when they encourage people 00:20:17.769 --> 00:20:20.337 to go and explore that expanse for themselves, 00:20:20.337 --> 00:20:24.490 and The Art of Problem Solving is among the few great places to actually do 00:20:24.490 --> 00:20:25.420 that exploration.